turning operations
DESCRIPTION
Turning Operations. L a t h e. Turning Operations. Machine Tool – LATHE Job (workpiece) – rotary motion Tool – linear motions “ Mother of Machine Tools “ Cylindrical and flat surfaces. Some Typical Lathe Jobs. Turning/Drilling/Grooving/ Threading/Knurling/Facing. The Lathe. The Lathe. - PowerPoint PPT PresentationTRANSCRIPT
Turning Operations
L a t h e
Turning Operations
• Machine Tool – LATHE • Job (workpiece) – rotary motion• Tool – linear motions
“Mother of Machine Tools “Cylindrical and flat surfaces
Some Typical Lathe JobsTurning/Drilling/Grooving/Threading/Knurling/Facing...
The Lathe
Head stock
Spindle
Feed rod
Bed
Compound restand slide(swivels)
CarriageApron
selector
Feed changegearbox
Spindlespeed
Lead screw
Cross slideDead center
Tool postGuide ways
Tailstock quillTailstock
Handle
The Lathe
Bed
Head StockTail Stock
CarriageFeed/Lead Screw
Main Parts
• Bed• Headstock• Feed and lead screws• Carriage• Tailstock
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Lathe Bed• Heavy, rugged casting• Made to support working parts of lathe• On top section are machined ways
– Guide and align major parts of lathe
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Lathe Bed
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Headstock• Clamped on left-hand end of bed• Headstock spindle
– Hollow cylindrical shaft supported by bearings• Provides drive through gears to work-holding
devices– Live center, faceplate, or chuck fitted to spindle nose
to hold and drive work• Driven by stepped pulley or transmission gears• Feed reverse lever
– Reverses rotation of feed rod and lead screw
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Headstock
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Headstock
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Back Gear arrangement
Headstock belt drive
Quick-Change Gearbox• Contains number of different-size gears• Provides feed rod and lead-screw with various
speeds for turning and thread-cutting operations– Feed rod advances carriage when automatic
feed lever engaged– Lead screw advances the carriage for thread-
cutting operations when split-nut lever engaged
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Quick-Change Gearbox
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Carriage• Used to move cutting tool along lathe bed• Consists of three main parts
– Saddle• H-shaped casting mounted on top of lathe ways,
provides means of mounting cross-slide and apron– Cross-slide– Apron
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Carriage
< Saddle
< Apron
Carriage
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Carriage
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Apron
• The apron attached to the front of the carriage, holds most of the control levers. These include the levers, which engage and reverse the feed lengthwise (Z-axis) or crosswise (X-axis) and the lever which engages the threading gears.
• The apron is fastened to the saddle, houses the gears and mechanisms required to move the carriage and cross-slide automatically.
• The apron hand wheel can be turned manually to move the carriage along the Lathe bed. This hand wheel is connected to a gear that meshes in a rack fastened to the Lathe bed.
• The automatic feed lever engages a clutch that provides the automatic feed to the carriage
Cross-slide• Mounted on top of saddle• Provides manual or automatic cross movement
for cutting tool• Compound rest (fitted on top of cross-slide)
– Used to support cutting tool– Swiveled to any angle for taper-turning– Has graduated collar that ensure accurate
cutting-tool settings (.001 in.) (also cross-slide)
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Cross-slide
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Top Slide (Compound slide)
• Fitted to top of Cross slide• Carries tool post and cutting tool• Can rotate to any angle• Is used to turn tapers
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Tailstock• Upper and lower tailstock castings• Adjusted for taper or parallel turning by two
screws set in base• Tailstock clamp locks tailstock in any position
along bed of lathe• Tailstock spindle has internal taper to receive
dead center– Provides support for right-hand end of work
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Tailstock
Supports long workpieces when machining.
60 degree rotating center point.
Drill Chuck
Turn the tailstock handwheel to advance the ram.
Tailstock
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Lead Screw and Feed Rod
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< Lead Screw
< Feed Rod
Types of Lathes
• Engine Lathe• Speed Lathe• Bench Lathe
• Tool Room Lathe• Special Purpose Lathe
• Gap Bed Lathe…
Size of LatheWorkpiece Length Swing
Size of Lathe ..Example: 300 - 1500 Lathe• Maximum Diameter of Workpiece that can
be machined = SWING (= 300 mm)
• Maximum Length of Workpiece that can be held between Centers (=1500 mm)
Workholding Devices
• Equipment used to hold– Workpiece – fixtures– Tool - jigs
Securely HOLD or Support while machining
Chucks
Three jaw Four Jaw
Wor
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Chucks• Used extensively for holding work for
lathe machining operations– Work large or unusual shape
• Most commonly used lathe chucks– Three-jaw universal– Four-jaw independent– Collet chuck
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Three-jaw Universal Chuck
• Holds round and hexagonal work• Grasps work quickly and accurate
within few thousandths/inch• Three jaws move simultaneously when
adjusted by chuck wrench– Caused by scroll plate into which all three
jaws fit• Two sets of jaw: outside chucking and
inside chucking
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Three-jaw Universal Chuck
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Three jaw self centering chuck
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Four-Jaw Independent Chuck
• Used to hold round, square, hexagonal, and irregularly shaped workpieces
• Has four jaws– Each can be adjusted independently by
chuck wrench• Jaws can be reversed to hold work by
inside diameter
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Four-Jaw Independent Chucks
Four-Jaw Independent Chucks
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• With the four jaw chuck, each jaw can be adjusted independently by rotation of the radially mounted threaded screws.
• Although accurate mounting of a workpiece can be time consuming, a four-jaw chuck is often necessary for non-cylindrical workpieces.
MandrelsWorkpiece (job) with a hole
Wor
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D
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Mandrels
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• Holds internally machined workpiece between centers so further machining operations are concentric with bore
• Several types, but most common• Plain mandrel• Expanding mandrel• Gang mandrel• Stub mandrel
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Mandrels to Hold Workpieces for Turning
Figure 23.8 Various types of mandrels to hold workpieces for turning. These mandrels usually are mounted between centers on a lathe. Note that in (a), both the cylindrical and the end faces of the workpiece can be machined, whereas in (b) and (c), only the cylindrical surfaces can be machined.
RestsW
orkh
oldi
ng
Dev
ices
.. JawsHingeWork Work Jaws
Lathe bed guideways
Carriage
Steady Rest Follower Rest
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Steadyrest• Used to support long work held in chuck
or between lathe centers– Prevent springing
• Located on and aligned by ways of the lathe
• Positioned at any point along lathe bed• Three jaws tipped with plastic, bronze or
rollers may be adjusted to support any work diameter with steadyrest capacity
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Steadyrest
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Follower Rest• Mounted on saddle• Travels with carriage to prevent work
from springing up and away from cutting tool– Cutting tool generally positioned just
ahead of follower rest– Provide smooth bearing surface for two
jaws of follower rest
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Follower Rest
Operating/Cutting Conditions
1. Cutting Speed v2. Feed f3. Depth of Cut d
Workpiece
Tool
Chip
Tool post
S peripheralspeed (m/min)
N (rev/min)
D
Operating Conditions
Workpiece
Tool
Chip
Tool post
S peripheralspeed (m/min)
N (rev/min)
D
NDSspeedperipheralD
rotation1intraveltoolrelative
Cutting Speed
The Peripheral Speed of Workpiece past the Cutting Tool
=Cutting SpeedOpe
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Cond
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m/min1000
NDv
D – Diameter (mm)N – Revolutions per Minute (rpm)
Feedf – the distance the tool advances for every
rotation of workpiece (mm/rev)
Ope
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Cond
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f Feed
DD 21
Depth of Cutperpendicular distance between machined
surface and uncut surface of the Workpiece d = (D1 – D2)/2 (mm)
Ope
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Cond
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d Depth of Cut
DD 21
3 Operating Conditions
Chip
Machined surface
Workpiece
Depth of cutToolChuck
N
Feed (f )
Cutting speed
Depth of cut (d)
Selection of ..
• Workpiece Material • Tool Material• Tool signature • Surface Finish• Accuracy • Capability of Machine Tool
Ope
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Cond
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Material Removal Rate
MRRVolume of material removed in one
revolution MRR = D d f mm3
• Job makes N revolutions/minMRR = D d f N (mm3/min)
• In terms of v MRR is given byMRR = 1000 v d f (mm3/min)O
pera
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MRR
dimensional consistency by substituting the units
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MRR: D d f N (mm)(mm)(mm/rev)(rev/min) = mm3/min
Operations on Lathe
• Turning• Facing• knurling• Grooving• Parting
• Chamfering• Taper turning• Drilling• Threading
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Turning
Cylindrical job
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Turning ..
Cylindrical job
Cutting speed
Chip
WorkpieceDepth of cut (d)
Depth of cutTool
FeedChuck
N Machined surface
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Turning ..
• Excess Material is removed to reduce Diameter
• Cutting Tool: Turning Tool
a depth of cut of 1 mm will reduce diameter by 2 mm
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FacingFlat Surface/Reduce length
Depth of cut
Feed
WorkpieceChuck
Cutting speed
Tool
dMachined Face
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Facing ..
• machine end of job Flat surfaceor to Reduce Length of Job
• Turning Tool• Feed: in direction perpendicular to
workpiece axis–Length of Tool Travel = radius of
workpiece• Depth of Cut: in direction parallel to
workpiece axisOpe
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Facing ..O
pera
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Eccentric Turning
Axis of job
Axis of lathe
Eccentric peg(to be turned)
4-jaw chuck
Cutting speed
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Knurling
• Produce rough textured surface– For Decorative and/or Functional Purpose
• Knurling Tool
A Forming ProcessMRR~0
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KnurlingO
pera
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s on
La
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.. Knurling toolTool post
Feed
Cutting speed
Movement for depth
Knurled surface
Knurling ..O
pera
tion
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Grooving• Produces a Groove on
workpiece• Shape of tool shape of groove• Carried out using Grooving Tool A form tool
• Also called Form Turning
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Grooving ..O
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.. Shape produced by form tool Groove
Grooving tool
Feed or depth of cutForm tool
Parting
• Cutting workpiece into Two• Similar to grooving• Parting Tool• Hogging – tool rides over – at slow feed• Coolant use
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Parting ..O
pera
tion
s on
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FeedParting tool
ChamferingO
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Chamfering toolFeed
Chamfer
Chamfering
Beveling sharp machined edges Similar to form turning Chamfering tool – 45° To
• Avoid Sharp Edges• Make Assembly Easier• Improve Aesthetics
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Taper Turning
• Taper:
CB
A L
D90° 2D1
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LDD
2tan 21
Taper Turning..
Methods • Form Tool• Swiveling Compound Rest• Taper Turning Attachment• Simultaneous Longitudinal and
Cross FeedsOpe
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ConicityL
DDK 21
Taper Turning ..By Form Tool
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TaperWorkpiece
Straight cutting edge
Direction of feed
Form tool
Taper Turning ,,By Compound Rest
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Face plate
Dog
Tail stock quill
Tail stock
Mandrel
Direction of feedCompound rest
SlideCompound rest Hand crank
Tool post & Tool holder
Cross slide
DrillingDrill – cutting tool – held in TS – feed from TS
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Feed
DrillQuill clamp moving
quill
Tail stock clamp
Tail stock
Process Sequence• How to make job from raw material 45 long
x 30 dia.?
20 dia
40
15
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Steps:•Operations
•Sequence•Tools•Process
Process Sequence .. Possible Sequences
• TURNING - FACING - KNURLING• TURNING - KNURLING - FACING• FACING - TURNING - KNURLING• FACING - KNURLING - TURNING• KNURLING - FACING - TURNING• KNURLING - TURNING – FACINGWhat is an Optimal Sequence?
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X
XXX
Machining Time
Turning Time• Job length Lj mm• Feed f mm/rev• Job speed N rpm• f N mm/min
min Nf
Lt jO
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Manufacturing Time
Manufacturing Time = Machining Time + Setup Time + Moving Time + Waiting Time
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ExampleA mild steel rod having 50 mm diameter and
500 mm length is to be turned on a lathe. Determine the machining time to reduce the rod to 45 mm in one pass when cutting speed is 30 m/min and a feed of 0.7 mm/rev is used.
Example
calculate the required spindle speed as: N = 191 rpm
m/min1000
NDv
Given data: D = 50 mm, Lj = 500 mm v = 30 m/min, f = 0.7 mm/rev
Substituting the values of v and D in
ExampleCan a machine has speed of 191
rpm?Machining time:
min Nf
Lt j
t = 500 / (0.7191) = 3.74 minutes
Example
• Determine the angle at which the compound rest would be swiveled for cutting a taper on a workpiece having a length of 150 mm and outside diameter 80 mm. The smallest diameter on the tapered end of the rod should be 50 mm and the required length of the tapered portion is 80 mm.
Example
• Given data: D1 = 80 mm, D2 = 50 mm, Lj = 80 mm (with usual notations)
tan = (80-50) / 280• or = 10.620• The compound rest should be swiveled
at 10.62o
Example
• A 150 mm long 12 mm diameter stainless steel rod is to be reduced in diameter to 10 mm by turning on a lathe in one pass. The spindle rotates at 500 rpm, and the tool is traveling at an axial speed of 200 mm/min. Calculate the cutting speed, material removal rate and the time required for machining the steel rod.
Example
• Given data: Lj = 150 mm, D1 = 12 mm, D2 = 10 mm, N = 500 rpm
• Using Equation (1)• v = 12500 / 1000• = 18.85 m/min.• depth of cut = d = (12 – 10)/2 = 1 mm
Example• feed rate = 200 mm/min, we get the feed f in
mm/rev by dividing feed rate by spindle rpm. That is
• f = 200/500 = 0.4 mm/rev• From Equation (4),• MRR = 3.142120.41500 = 7538.4
mm3/min• from Equation (8),• t = 150/(0.4500) = 0.75 min.
Example
• Calculate the time required to machine a workpiece 170 mm long, 60 mm diameter to 165 mm long 50 mm diameter. The workpiece rotates at 440 rpm, feed is 0.3 mm/rev and maximum depth of cut is 2 mm. Assume total approach and overtravel distance as 5 mm for turning operation.
Example
• Given data: Lj = 170 mm, D1 = 60 mm, D2 = 50 mm, N = 440 rpm, f = 0.3 mm/rev, d= 2 mm,
• How to calculate the machining time when there is more than one operation?
Example
• Time for Turning:• Total length of tool travel = job length + length of approach
and overtravel• L = 170 + 5 = 175 mm• Required depth to be cut = (60 – 50)/2 = 5 mm• Since maximum depth of cut is 2 mm, 5 mm cannot be cut
in one pass. Therefore, we calculate number of cuts or passes required.
• Number of cuts required = 5/2 = 2.5 or 3 (since cuts cannot be a fraction)
• Machining time for one cut = L / (fN)• Total turning time = [L / (fN)] Number of cuts• = [175/(0.3440)] 3= 3.97
min.
Example
• Time for facing:• Now, the diameter of the job is reduced
to 50 mm. Recall that in case of facing operations, length of tool travel is equal to half the diameter of the job. That is, l = 25 mm. Substituting in equation 8, we get
• t = 25/(0.3440)• = 0.18 min.
Example
• Total time:• Total time for machining = Time
for Turning + Time for Facing• = 3.97 + 0.18• = 4.15 min. • The reader should find out the total
machining time if first facing is done.
Example
• From a raw material of 100 mm length and 10 mm diameter, a component having length 100 mm and diameter 8 mm is to be produced using a cutting speed of 31.41 m/min and a feed rate of 0.7 mm/revolution. How many times we have to resharpen or regrind, if 1000 work-pieces are to be produced. In the taylor’s expression use constants as n = 1.2 and C = 180
Example
• Given D =10 mm , N = 1000 rpm, v = 31.41 m/minute
• From Taylor’s tool life expression, we havevT n = C
• Substituting the values we get,• (31.40)(T)1.2 = 180• or T = 4.28 min
Example
• Machining time/piece = L / (fN)• = 100 / (0.71000)• = 0.142 minute.• Machining time for 1000 work-pieces =
1000 0.142 = 142.86 min• Number of resharpenings = 142.86/ 4.28 • = 33.37 or 33 resharpenings
Example
• 6: While turning a carbon steel cylinder bar of length 3 m and diameter 0.2 m at a feed rate of 0.5 mm/revolution with an HSS tool, one of the two available cutting speeds is to be selected. These two cutting speeds are 100 m/min and 57 m/min. The tool life corresponding to the speed of 100 m/min is known to be 16 minutes with n=0.5. The cost of machining time, setup time and unproductive time together is Rs.1/sec. The cost of one tool re-sharpening is Rs.20.
• Which of the above two cutting speeds should be selected from the point of view of the total cost of producing this part? Prove your argument.
Example
• Given T1 = 16 minute, v1 = 100 m/minute, v2 = 57 m/minute, D = 200mm, l = 300 mm, f = 0.5 mm/rev
• Consider Speed of 100 m/minute• N1 = (1000 v) / ( D) = (1000100) /
(200) = 159.2 rpm• t1 = l / (fN) = 3000 / (0.5 159.2) = 37.7
minute• Tool life corresponding to speed of 100
m/minute is 16 minute.• Number of resharpening required = 37.7 / 16 =
2.35•• or number of resharpenings = 2
Example
• Total cost =• Machining cost + Cost of resharpening
Number of resharpening• = 37.7601+ 202 • = Rs.2302
Example
• Consider Speed of 57 m/minute• Using Taylor’s expression T2 = T1
(v1 / v2)2 with usual notations• = 16 (100/57)2 = 49 minute• Repeating the same procedure we get t2 =
66 minute, number of reshparpening=1 and total cost = Rs. 3980.
•• The cost is less when speed = 100
m/minute. Hence, select 100 m/minute.
Example
• Write the process sequence to be used for manufacturing the component
from raw material of 175 mm length and 60 mm diameter
Example
40 Dia50
50 40
20
50
20 Dia
Threading
Example
• To write the process sequence, first list the operations to be performed. The raw material is having size of 175 mm length and 60 mm diameter. The component shown in Figure 5.23 is having major diameter of 50 mm, step diameter of 40 mm, groove of 20 mm and threading for a length of 50 mm. The total length of job is 160 mm. Hence, the list of operations to be carried out on the job are turning, facing, thread cutting, grooving and step turning
Example• A possible sequence for producing the
component would be:• Turning (reducing completely to 50 mm)• Facing (to reduce the length to 160 mm)• Step turning (reducing from 50 mm to 40
mm)• Thread cutting.• Grooving